Problem: A box contains $3$ red marbles, $4$ green marbles, and $3$ blue marbles. If we choose a marble, then another marble without putting the first one back in the box, what is the probability that the first marble will be blue and the second will be blue as well? Write your answer as a simplified fraction.
The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened In this case, event A is picking a blue marble and leaving it out. Event B is picking another blue marble. Let's take the events one at at time. What is the probability that the first marble chosen will be blue? There are $3$ blue marbles, and $10$ total, so the probability we will pick a blue marble is $\dfrac{3} {10}$ After we take out the first marble, we don't put it back in, so there are only $9$ marbles left. Also, we've taken out one of the blue marbles, so there are only $2$ left altogether. So, the probability of picking another blue marble after taking out a blue marble is $\dfrac{2} {9}$ Therefore, the probability of picking a blue marble, then another blue marble is $\dfrac{3}{10} \cdot \dfrac{2}{9} = \dfrac{1}{15}$